4) The total number of outcomes for two dice being tossed is (6)(6)=36. The number of outcomes that the first die shows a 3 and the second die shows NOT a 3 is five: (3,1), (3,2), (3,4), (3,5) and (3,6). The same number of outcomes – five, is for the second die to show the 3, and the first die to be different from 3. Then, the probability of both dice to show exactly one 3 is: P(exactly one 3) =P(3 on 1st die, not 3 on 2nd die) + P(not 3 on 1st die, 3 on 2nd die =5/36+5/36=10/36.There is the only one event to get exactly two 3s – (3,3); the corresponding probability is P(two 3s) = 1/36.The sought for probability is the sum of the probability of showing exactly one 3 and the probability of showing exactly two 3s’ (in two tosses):P(at least one 3) = P(exactly one 3) + P(two 3) = 10/36 + 1/36 = 11/36

5) Let us trace the possibility for the marble to be black. The probability of black marble taken from the red bag is 4/7. If the coin shows the heads (P(H)=1/2), then the probability of both events isP(H)P(Black marble from red bag)=(1/2)(4/7)=2/7Consequently, the probability of black marble taken from the yellow bag is 2/7. If the coin shows the tails (P(T)=1/2), then the probability of both events is: P(T)P(Black marble from yellow bag)=(1/2)(2/7)=1/7. Finally, P(Black marble)= 2/7+1/7=3/7

6) The probability of two heads followed by two tails (in this special order) is P(HHTT)=(1/2)^2(1/2)^2 =1/16. There are 6 possible arrangements of two heads and two tails: HHTT, HTHT, HTTH, THHT, THTH and TTHH with the same probability of 1/6. The probability of two heads and two tails without specifying the order is: P=6(1/16)=3/8